glamrockqueen7

2021-08-20

To calculate: The product of $\left[x-\left(3-\sqrt{5}\right)\right]\left[x-\left(3+\sqrt{5}\right)\right]$

Demi-Leigh Barrera

Step 1
Formula used:
$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$
Associative property, $\left(a+b\right)+c=a+\left(b+c\right)$
Step 2
If P (x) represent the given expression, then
$P\left(x\right)=\left[x-\left(3-\sqrt{5}\right)\right]\left[x-\left(3+\sqrt{5}\right)\right]$
$P\left(x\right)=\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)$
Use associative property of algebraic expressions,
Associative property is written as,
$\left(a+b\right)+c=a+\left(b+c\right)$
This property modifies the expression to,
$P\left(x\right)=\left(\left(x-3\right)+\sqrt{5}\right)\left(\left(x-3\right)-\sqrt{5}\right)$
Apply arithmetic rule:
$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
Here,
Hence,
$P\left(x\right)={\left(x-3\right)}^{2}-{\left(\sqrt{5}\right)}^{2}Z$
Apply arithmetic rule:
${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$
Here,
Hence,
$P\left(x\right)={\left(x-3\right)}^{2}-{\left(\sqrt{5}\right)}^{2}$
$={x}^{2}-6x+9-5$
$={x}^{2}-6x+4$
Hence, the product of $\left[x-\left(3-\sqrt{5}\right)\right]\left[x-\left(3+\sqrt{5}\right)\right]$ is ${x}^{2}-6x+4$

Jeffrey Jordon